Denoting the order of $g$ by $o(g)$, you can show that for any decreasing function $f$ the following inequality holds $$\sum_{g\in G}f(o(g))\geq \sum_{g\in \mathbb Z/n\mathbb Z}f(o(g)).$$ This is because one can actually construct a bijection $\sigma:G\to\mathbb Z/n\mathbb Z$ which satisfies $$o(\sigma(g))\geq o(g)$$ for all $g\in G$. The main ingredient is a classical theorem of Frobenius saying that when $k$ divides the order of a group, the number of elements of order dividing $k$ in a finite group, is divisible by $k$, then proceed by induction. An application of this exact idea is for example problem 10775 on the American Math Monthly. For your question we just need $f(x)=-\log x$.
Denoting the order of $g$ by $o(g)$, you can show that for any decreasing function $f$ the following inequality holds $$\sum_{g\in G}f(o(g))\geq \sum_{g\in \mathbb Z/n\mathbb Z}f(o(g)).$$ This is because one can actually construct a bijection $\sigma:G\to\mathbb Z/n\mathbb Z$ which satisfies $$o(\sigma(g))\geq o(g)$$ for all $g\in G$. The main ingredient is a classical theorem of Frobenius saying that the number of elements of order dividing $k$ in a finite group, is divisible by $k$, then proceed by induction. An application of this exact idea is for example problem 10775 on the American Math Monthly. For your question we just need $f(x)=-\log x$.
Denoting the order of $g$ by $o(g)$, you can show that for any decreasing function $f$ the following inequality holds $$\sum_{g\in G}f(o(g))\geq \sum_{g\in \mathbb Z/n\mathbb Z}f(o(g)).$$ This is because one can actually construct a bijection $\sigma:G\to\mathbb Z/n\mathbb Z$ which satisfies $$o(\sigma(g))\geq o(g)$$ for all $g\in G$. The main ingredient is a classical theorem of Frobenius saying that the number of elements of order dividing $k$ in a finite group, is divisible by $k$, then proceed by induction. An application of this exact idea is for example problem 10775 on the American Math Monthly. For your question we just need $f(x)=-\log x$.